Standard proof that the set of sigularities is closed

Question

I am trying to prove that the set of singular points of an affine variety is closed.

Suppose $X\subset \mathbb{A}^n$. As every affine tangent space $T_x$ is embedded in this $\mathbb{A}^n$ we consider the tangent bundle $TX$ as a subvariety of the product $X\times\mathbb{A}^n$. After that we introduce the projection $\pi:TX\to X$, which maps each pair $(x,v)(\in X\times T_xX)$ to $x$ (it means that the fibres of $\pi$ are exactly the tangent spaces). There are the so called fibre dimension theorems which state that given a dominant morphism of irreducible varieties $f: X\to Y$ then $$\forall y\in Y\space \mathrm{dim}(f^{-1}(y))\ge \mathrm{dim}(X)-\mathrm{dim}(Y))$$ and there exists an open subset $U\subset Y$ such that for all $y\in U$ $$\mathrm{dim}(f^{-1}(y))=\mathrm{dim}(X)-\mathrm{dim}(Y).$$ How can I use it to prove the original statement?

Hm... A point $x\in X$ is singular if and only if $\mathrm{dim}(T_xX)>\mathrm{dim}(X)$. Yes, it is. That is a definition. Also, for all $k\in \mathbb{N}$ the set $\{y\in Y:\space \mathrm{dim}(f^{-1}(y)\ge k)\}$ is closed. That is, the set of all $x\in X$ such that $\mathrm{dim}(T_xX)\ge \mathrm{dim}(X)$ is closed. But we need the strict inequality.

I am so confused, could you help please?

Answer 1

I don't see how to apply the first set of statements directly, but they do imply this semicontinuity of the fiber dimension that you mention at the end, and that tells you that the set of $x$ such that $\dim T_xX \geq \dim X + 1$ is closed; since (finite) dimensions are integers this does what you want.

This is a cool use of these facts but I want to add that what I consider the "standard" proof uses less technology: a point is singular iff the Jacobian matrix has rank $< n - \dim X$ there, so you write down the minors of size $n - \dim X$ and demand that they all vanish.